hw05_solutions - AA311 Atmospheric Flight Mechanics Autumn...

AA311 - Atmospheric Flight MechanicsAutumn 2010University of WashingtonHomework 5 SolutionsProblem 1Part a. For standard sea level conditionsT¶=288.16Kr¶=1.225kgëm3p¶=1.01325ä105Nëm2Tinfty=288.16;ρinfty=1.225;pinfty=1.01325∗105;We can first calculate the Mach numberM¶=V¶awherea=gR T¶Hspeed of soundLγ =1.4;R=287;Vinfty=50;a=γR TinftyMinfty=Vinftya340.2690.146943So we see that the speed of sound is 340 m/s and M¶=0.14 which is slow enough so we do not need to worry about compressib-lity. In this case, we can calculate the low mach number coefficient of pressure asCp,0=p-p¶Å12r¶V¶2wherep=9.5ä104Nëm2AA311 - Atmospheric Flight MechanicsChristopher LumPrinted by Mathematica for Students

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p=9.5∗104;Cp0=p−pinftyfractionbar12ρinfty∗Vinfty2−4.13061So we haveCp,0= -4.13Part b. If the velocity is increased to V¶=150mês, we can calculate the Mach numberVinfty=150;Minfty=Vinftya0.440828So we see that the Mach number is high enough that we need to apply compressiblity effects. The low speed coefficient ofpressure is given asCp,0=p-p¶Å12r¶V¶2Cp0=p−pinftyfractionbar12ρinfty∗Vinfty2−0.458957We can now apply the Prandtl-Glauert rule to obtain the actual coefficient of pressureCp=Cp,01-M¶2Cp=Cp01−Minfty2−0.511321So we haveCp= -0.511Clear@Cp, Cp0, Minfty, Vinfty, Cp0, p, Minfty, a, Vinfty, R,γ, pinfty,ρinfty, TinftyD2hw05_solutions.nbPrinted by Mathematica for Students

Problem 2Part a. We are interested in the pressure coefficient ofCp=p-p¶Å12r¶V¶2(Eq.1.1)If we assume that the flow is incompressible, we can apply Bernoulli's equation.p+ Å12rV2=p¶+ Å12r¶V¶2p-p¶= Å12Ir¶V¶2- rV2M(Eq.1.2)Substituting Eq.1.2 into Eq.1.1 yieldsCp=Å12Ir¶V¶2- rV2MÅ12r¶V¶2=r¶V¶2- rV2r¶V¶2=1-rV2r¶V¶2recall: incompressible so r = r¶Cp=1-V2V¶2V=195;Vinfty=160;Cp=1−V2Vinfty2êêN−0.485352So we haveCp= -0.485This is not a function of density since it only depends on the ratio of velocities, so it would not matter where the wind tunnel islocatedClear@Cp, Vinfty, VDhw05_solutions.nb 3AA311 - Atmospheric Flight MechanicsChristopher LumPrinted by Mathematica for Students

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Problem 3From Appendix B, we find some free stream conditions ofT¶=483.04oRp¶=1.4556ä103lbfëft2r¶=1.7556ä10-3slugëft3Tinfty=483.04;pinfty=1.4556∗103;ρinfty=1.7556∗10−3;We can calculate the speed of sounda=gR T¶whereg =1.4R=1716lbf ftslugRγ =1.4;R=1716;a=γR Tinfty1077.24So we can find the free stream velocityV¶=M¶aMinfty=0.205;Vinfty=Minfty∗a220.835So we haveV¶=220.8 ftêsWe can calculate the Reynold's number. To do this, we need to first find m¶. From Figure 4.34, we see that forT¶=483.04oR=268.35oKm¶=1.65ä10-5kgm s=1.65ä10-5kgm sÿÀ2.20462262 lbm1 kgÀÿÀ1 slug32.2 lbmÀÿ…1m3.2808399 ft…4hw05_solutions.nbPrinted by Mathematica for Students

μinfty=1.65∗10−5∗2.204622621∗132.2∗13.28083993.44332×10−7So we havem¶=3.44ä10-7slugfts

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